# What is the correct definition of hydration enthalpy and why is it always negative?

The first definition of hydration enthalpy I found on the net is:

The hydration enthalpy is the enthalpy change when $1$ mole of gaseous ions dissolve in sufficient water to give an infinitely dilute solution. Hydration enthalpies are always negative.

Another definition I found in a book is:

Hydration enthalpy is the energy released when $1$ mole of a compound is dissolved in sufficient water to give an infinitely dilute solution.

Now I have three questions:

• Which of the two definitions is the correct one? Why?

• Why exactly are hydration enthalpies always negative? Suppose I have hydrated copper(II) sulphate which is mostly stable as $\ce{CuSO4.5H2O}$. I convert it to $\ce{CuSO4.6H2O}$ by adding more water forcefully. Shouldn't the hydration enthalpy be positive in this case as it will make the final product less stable than the initial one?

• Which definition of hydration enthalpy applies for $\ce{CuSO4}$? I do not think $\ce{CuSO4.5H2O}$ is "infinitely" dilute by any extension of logic. Please explain this if possible.

P.S: I am a novice in thermochemistry. It would be helpful if you could keep the explanations as simple as possible.

• I'd just like to note that both definitions are commonly called "enthalpy of hydration", so if you see it in a book, make sure to check the context and figure out which one they're referring to. It's very common to talk about the "heat of hydration of so-and-so ion", which refers to the first of your two definitions. Oct 24 '16 at 19:21
• Could you please answer "Suppose I have hydrated copper(II) sulphate which is mostly stable as $\ce{CuSO4.5H2O}$. I convert it to $\ce{CuSO4.6H2O}$ by adding more water forcefully. Shouldn't the hydration enthalpy be positive in this case as it will make the final product less stable than the initial one?" @orthocresol
– user14857
Oct 25 '16 at 3:25

Hydration is defined as the following process (with any salt but using copper(II) sulphate as an example):

$$\ce{CuSO4 + 5 H2O -> CuSO4 . 5 H2O (s)}\tag{1}$$

This reaction will only happen if the associated enthalpy $\Delta_\mathrm{hyd} H^0$ — or more precisely, the associated Gibbs free energy $\Delta_\mathrm{hyd} G^0$ is negative. For some compounds, that is not the case; for example sodium chloride does not form hydrates: reaction (2) does not occur spontaneously.

$$\ce{NaCl + n H2O} \nrightarrow \ce{NaCl . n H2O}\tag{2}$$

Thus, we cannot measure a hydration enthalpy. If we can measure it, the process must be spontaneous and thus the enthalpy negative. Thus, all measurable hydration enthalpies are negative.

Sometimes, salts can form multiple hydrates. However, not every hydrate is always possible. Hydration enthalpies only exist for those hydrates which are possible.

Your definitions are basically identical only that they do not measure hydration enthalpies but solvation enthalpies. That is the process as shown in equation (3).

$$\ce{CuSO4 (+ H2O) -> Cu^2+ (aq) + SO4^2- (aq) (+ H2O)}\tag{3}$$

For that process, the enthalpy does not have to be negative. In fact, quite a few ionic compounds dissolve endothermicly, e.g. $\ce{NH4Cl}$.

• Even the process $\ce{Na^+(g) \rightarrow Na^+(l)}$ is called hydration. Is the enthalpy for it always negative?
– user14857
Oct 25 '16 at 3:34
• FYI "Hydration energy (also hydration enthalpy) is the amount of energy released when one mole of ions undergo hydration which is a special case of solvation. It is a special case of dissolution energy, with the solvent being water." So the last example you gave falls under hydration only. See en.wikipedia.org/wiki/Hydration_energy
– user14857
Oct 25 '16 at 3:35
• I don't agree with "If we can measure it, the process must be spontaneous and thus the enthalpy negative." Spontaneous is about Gibbs energy, not enthalpy. I guess the entropy is expected to be negative (constraining water molecules to the crystal lattice) but unless you say that explicitly, the enthalpy could be positive or negative for a spontaneous reaction or process. Dec 6 '21 at 17:09
• @KarstenTheis Of course, it's not easy to think oneself back into a five year old answer ... However, I think I wrote that because starting material (ionic compound) and product (hydrate of the ionic compound) are both solids and somehow I recall that in those reactions the entropy term would become zero and only the enthalpy ends up determining $\Delta G$ and thus the spontaneity? It's been a long time and I could have drawn wrong conclusions from prior knowledge.
– Jan
Dec 8 '21 at 10:17

Definition 2 sounds like the definition for enthalpy of solution.

In that case, both are correct: they measure different things.

Hydration enthalpies are always negative because hydrating an ion would always stabilize it. For hydration enthalpies, note that the waters of hydration are absent: you are hydrating the gas phase ions.

You can also measure both for $\ce{CuSO4}$. There is no hydration enthalpy for the hydrate.

• No, $\ce{CuSO4}$ does not have a hydration enthalpy. Look at the definition again. You need to dissolve the component ions to infinite dilution. That preludes the ability to "hydrate" the hydrate.
– Zhe
Oct 24 '16 at 22:39

To me both definitions are correct. The complex cation hexaaquacopper(II) cannot undergo further hydration and therefore it is infinitely dilute given that all 6 ligands have been provided. From the definition, since hydration enthalpy is a measure of energy released from a mole of compound then the final compound possesses less energy than the initial one. enthalpy final minus enthalpy initial in this case will always be negative. in terms of stability the pentaaquacopper(II) complex should be unstable because 5 is not a common coordination number for the copper metal ion so the complex should be less stable and more energetic than the hydrated one.